Page 147 - FUNDAMENTALS OF COMPUTER

NPP

NPP Number System, Boolean Algebra and Logic Circuits 147

(a) .75 (b) .84
Solution:
(a) Product Integer Part
.75 × 2 = 1.50 1
.50 × 2 = 1.00 1

00
Now read the Integer Parts downward and A~ nyUmªH$m| D$na go ZrMo n‹T>mo Ÿ& AV… (.11) =
2
put a leading point. Thus : (.11) = (.75) 10 (.75) 10
2
(b) Apply the same procedure : (b) Cnamoº$ {d{Y AZwgma
Product Integer Part
.84 × 2 = 1.68 1
.68 × 2 = 1.36 1
.36 × 2 = 0.72 0
.72 × 2 = 1.44 1
.44 × 2 = 0.88 0
.88 × 2 = 1.76 1
.76 × 2 = 1.52 1
It is clear from the example that the division Bg CXmhaU go ñnîQ> h¡ {H$ AnyUmªH$m| H$s pñW{V _|
may terminate after a long time or it may not à{H«$`m ~hþV Xoa VH$ Mb gH$Vr h¡ `m Vmo BgH$m A§V Zht
terminate. Here we can approximate the result àmá {H$`m Om gH$Vm h¡ Ÿ& Bg pñW{V _| h_| gÝZrH$Q>Z
up to few places of binary point. Suppose we
want to approximate the result up to four H$aZm hmoJm Ÿ& ~mBZar _| _mZm {H$ {g\©$ Mma A§H$m| VH$ hr
places of binary point, the result can be CÎma {bIZm h¡, V~ h_ nyUmªH$ H$mo D$na go ZrMo n‹T>H$a
obtained by reading the integer parts {bI gH$Vo h¢ {H$ …
downward.
(.1101) = (.84) 10
2
If a mixed number is given containing both `{X EH$ {_{lV Xe_bd g§»`m Xr JB© hmo Vmo g§nyU©
Integer and fractional parts, we will have to g_ñ`m Xmo {hñgm| _| {d^m{OV H$aZm hmoJm Ÿ& nyUmªH$ H$m
divide the whole problem in two parts; integer n[adV©Z AbJ VWm AnyUmªH$ H$m n[adV©Z AbJ H$aZm
and fractional. Consider the problem shown in
the example below: hmoJm Ÿ& ZrMo {bIr g_ñ`m H$mo g_Pmo…
Problem 3.7 àíZ 3.7
Find the Binary equivalent for the decimal Xe_bd g§»`m 24.36 hoVw ~mBZar g§»`m àmá
number 24.36. H$amoŸ&

Solution: hc:
The given number is 24.36. The integer Xr JB© {_{lV g§»`m 24.36 h¡ Ÿ& Bg_| 24 nyUmªH$
part is 24. Its binary equivalent can be obtained h¡Ÿ& BgH$m ~mBZar Vwë` Bg àH$ma go àmá {H$`m Om
as below:
gH$Vm h¡Ÿ&